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3 years ago

Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.

1 answer:

5 0

The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%

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The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that

Sladkaya [172]

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = $1 \ atom \times (\dfrac{1 \ mole}{ 6.022 \times 10^{23} \ atoms}) \times ( \dfrac{4.003 \ grams}{ 1 \ mole})$

$=6.647 \times 10^{-24} \ grams$

The volume can be determined as folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = $\dfrac{0.10}{2}$ = 0.05 nm

Converting it into cm, we have:

$0.05 nm \times \dfrac{10^{-9} \ meters}{ 1 nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}$

$=5 \times 10^{-9} \ cm$

Assuming that it is a sphere, the volume of a sphere is

= $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi \times (5\times 10^{-9})^3$

= $5.236 \times 10^{-25} \ cm^3$

Finally, the density can be calcuated by using the formula :

$Density = \dfrac{mass}{volume}$

$D = \dfrac{6.647 \times 10^{-24} \ grams }{ 5.236 \times 10^{-25} \ cm^3}$

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

4 0

3 years ago

Read 2 more answers

Write the following number using Scientific notation : 13,215,296.50 *

ludmilkaskok [199]

Answer:

13.22x10^7

Explanation:

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3 years ago

How would an increase in carbon dioxide concentration in the atmosphere change the green house effect? A. Carbon dioxide would c

Elis [28]

Answer:

Explanation:

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3 years ago

At the beginning of an experiment, a scientist has 176 grams of radioactive goo. After 165 minutes, her sample has decayed to 5.

Paul [167]

The half-life equation is written as:

An = Aoe^-kt

We use this equation for the solution. We do as follows:

5.5 = 176e^-k(165)
k = 0.02
What is the half-life of the goo in minutes?

0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE

Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?

G(t) = 176e^-0.02t

How many grams of goo will remain after 50 minutes?

G(t) = 176e^-0.02(50) = 64.75 g

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3 years ago

What was the velocity over the entire trip? A. -2 B. -1 C. 0 D. 1 E. 2

zhannawk [14.2K]

You’re right its A -2

5 0

2 years ago